# Indices up and down

I found notation in the form
##\vec{e}_{\alpha} \cdot \vec{x}=x^{\alpha}##
and also
## \langle \vec{e}^{* \alpha}, \vec{x} \rangle =x^{\alpha} ##.
If I understand this ##\vec{e}_{\alpha}## is unit vector in direct space, and ##\vec{e}^{* \alpha}## is unit vector in dual space, and the bracket ##\langle, \rangle## is like dual vector on vector on direct space. But why result is ##x^{\alpha}##?

martinbn
The ##x^\alpha## is the ##\alpha##-component of ##\vec{x}## in the basis ##\{\vec{e}_\alpha\}##. So ##\vec{x}=x^\alpha\vec{e}_\alpha##, the rest is the definition of the inner product and what a dual basis is.

Dale
Orodruin
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I found notation in the form
##\vec{e}_{\alpha} \cdot \vec{x}=x^{\alpha}##
This is true only in an orthonormal basis. In general,
$$\vec e_\alpha \cdot \vec x = \vec e_\alpha \cdot x^\beta \vec e_\beta = g_{\alpha\beta} x^\beta.$$

Pencilvester
Ok. Tnx. I am liitle bit confused with notation.
$$\frac{\partial}{\partial x^{\mu}}=\partial x_{\mu}$$?
So if I write
$$\frac{\partial}{\partial x^{\mu}} x^{\nu}=\partial x_{\mu}x^{\nu}=\delta_{\mu}^{\nu}$$.
So if I have for example
$$\frac{\partial}{\partial x_{\mu}}=\partial x^{\mu}?$$
Is it correct?

Orodruin
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Ok. Tnx. I am liitle bit confused with notation.
$$\frac{\partial}{\partial x^{\mu}}=\partial x_{\mu}$$?
This is not standard notation. The standard notation is
$$\frac{\partial}{\partial x^\mu} = \partial_\mu.$$
So if I write
$$\frac{\partial}{\partial x^{\mu}} x^{\nu}=\partial x_{\mu}x^{\nu}=\delta_{\mu}^{\nu}$$.
It is not so much a question about writing as it is about performing the derivative, but yes, ##\partial_\mu x^\nu = \delta^\nu_\mu##.
So if I have for example
$$\frac{\partial}{\partial x_{\mu}}=\partial x^{\mu}?$$
Is it correct?
This would be ##\partial/\partial x_\mu = \partial^\mu##.

Side-note: shoudn't this be I level and not B?

This is not standard notation. The standard notation is
$$\frac{\partial}{\partial x^\mu} = \partial_\mu.$$
Question: is that because

$$\frac{\partial}{\partial x^\mu}$$

is an operator, while

$$\partial x_\mu.$$

would be a specific operation on xμ?

Trying to learn these terms. Thanks.

Nugatory
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Side-note: shoudn't this be I level and not B?
Yes, good point. Fixed. (And you can report these instead of asking in thread - generally gets a faster response)

Sorcerer
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while

$$\partial x_\mu.$$

would be a specific operation on xμ?
This notation does not make sense to me at all. You would first have to identify what you mean by just ##\partial## without the index.

This notation does not make sense to me at all. You would first have to identify what you mean by just ##\partial## without the index.
Well, I was seeing it as just a differential of xμ. Is there such thing as just a partial differential, in the sense of say, dx, dy, dz?

But my other question, is that right? It's an operator?

Orodruin
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That operator is completely different and typically denoted with d, not ##\partial##. The one-forms ##dx^\mu## form a basis of the dual space.

PeroK
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Question: is that because

$$\frac{\partial}{\partial x^\mu}$$

is an operator, while

$$\partial x_\mu.$$

would be a specific operation on xμ?

Trying to learn these terms. Thanks.

To some extent you have to learn the specific notation and accept it. You can't necessarily deduce what a shorthand means from some absolute first principles.

The standard notation is
$$\frac{\partial}{\partial x^\mu} = \partial_\mu.$$

You have to accept this. It's just convention to drop the ##x## on the RHS. You're not going to find a deep mathematical reason why there's an ##x## in one and not the other. It's just the way people choose to write things. There's a certain logic in it, of course, but it could have been different.

stevendaryl
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This would be ##\partial/\partial x_\mu = \partial^\mu##.

Does ##x_\mu## have a standard meaning? ##x^\mu## means a coordinate, usually, but what is ##x_\mu##?

Orodruin
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